
When a green solid P was heated, it turned into a black solid Q and a colourless gas X was given off. On passing gas X overheated carbon, another colourless gas Y was formed. Gas Y burned in oxygen to reform gas X.
Identify P and Q.
(A) $CuC{{O}_{3}},CuO$
(B) $C{{u}_{2}}{{(C{{O}_{3}})}_{3}},C{{u}_{2}}O$
(C) $CaC{{O}_{3}},CuO$
(D) $CuCO,C{{O}_{2}}$
Answer
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In the reaction, on applying heat, it causes decomposition of the reactant with the release of gas, which can be detected by passing it through limewater, turning it milky. This gas can again be produced by, first combining with the heated carbon forming poisonous gas, which reacts to oxygen to give back the gas.
Complete step by step solution:
The green coloured solid P is the copper carbonate or the cupric carbonate salt, with the copper in $(+2)$ oxidation state.
On heating the copper carbonate, it breaks down into two substances, that is, copper (II) oxide, which is black solid (Q) and colourless carbon-dioxide gas (X). It is an endothermic reaction.
Due to this thermal decomposition of the copper carbonate with the release of gas, there is a loss in the mass of the copper oxide formed compared to the mass of carbonate initially taken. The reaction is as follows:
\[CuC{{O}_{3}}(s)\xrightarrow{\Delta }CuO(s)+C{{O}_{2}}(g)\] ------- (a)
Further on passing the carbon-dioxide gas (X) over the heated carbon, we get the colourless carbon-monoxide gas (Y). This combination reaction is given as follows:
\[C{{O}_{2}}(g)+C\to 2CO(g)\]
When the carbon-monoxide is burned in oxygen, it forms back the carbon-dioxide gas, with the release of heat. Thus, an exothermic reaction, which takes place as follows:
\[2CO(g)+{{O}_{2}}(g)\to 2C{{O}_{2}}(g)\]
Therefore, in the given reaction, substance P and Q is option (A)-$CuC{{O}_{3}}, CuO$ respectively.
Note: The carbon-dioxide released in reaction (a) can be detected by passing the gas through lime water, which turns milky.
This thermal decomposition reaction can also be used to determine the carbonates as the Group 2 alkaline metals readily decompose into metal oxides, whereas, the Group 1 alkali metals do not break, as being highly stable in nature.
Complete step by step solution:
The green coloured solid P is the copper carbonate or the cupric carbonate salt, with the copper in $(+2)$ oxidation state.
On heating the copper carbonate, it breaks down into two substances, that is, copper (II) oxide, which is black solid (Q) and colourless carbon-dioxide gas (X). It is an endothermic reaction.
Due to this thermal decomposition of the copper carbonate with the release of gas, there is a loss in the mass of the copper oxide formed compared to the mass of carbonate initially taken. The reaction is as follows:
\[CuC{{O}_{3}}(s)\xrightarrow{\Delta }CuO(s)+C{{O}_{2}}(g)\] ------- (a)
Further on passing the carbon-dioxide gas (X) over the heated carbon, we get the colourless carbon-monoxide gas (Y). This combination reaction is given as follows:
\[C{{O}_{2}}(g)+C\to 2CO(g)\]
When the carbon-monoxide is burned in oxygen, it forms back the carbon-dioxide gas, with the release of heat. Thus, an exothermic reaction, which takes place as follows:
\[2CO(g)+{{O}_{2}}(g)\to 2C{{O}_{2}}(g)\]
Therefore, in the given reaction, substance P and Q is option (A)-$CuC{{O}_{3}}, CuO$ respectively.
Note: The carbon-dioxide released in reaction (a) can be detected by passing the gas through lime water, which turns milky.
This thermal decomposition reaction can also be used to determine the carbonates as the Group 2 alkaline metals readily decompose into metal oxides, whereas, the Group 1 alkali metals do not break, as being highly stable in nature.
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